3.855 \(\int \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=147 \[ -\frac {(B+i A) (c-i d) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(-B+i A) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {A (d+i c)+B (c+3 i d)}{2 a f \sqrt {a+i a \tan (e+f x)}} \]

[Out]

-1/4*(I*A+B)*(c-I*d)*arctanh(1/2*(a+I*a*tan(f*x+e))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/f*2^(1/2)+1/2*(B*(c+3*I*d)+
A*(I*c+d))/a/f/(a+I*a*tan(f*x+e))^(1/2)+1/3*(I*A-B)*(c+I*d)/f/(a+I*a*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.30, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3590, 3526, 3480, 206} \[ -\frac {(B+i A) (c-i d) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(-B+i A) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {A (d+i c)+B (c+3 i d)}{2 a f \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c + d*Tan[e + f*x]))/(a + I*a*Tan[e + f*x])^(3/2),x]

[Out]

-((I*A + B)*(c - I*d)*ArcTanh[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(2*Sqrt[2]*a^(3/2)*f) + ((I*A - B
)*(c + I*d))/(3*f*(a + I*a*Tan[e + f*x])^(3/2)) + (B*(c + (3*I)*d) + A*(I*c + d))/(2*a*f*Sqrt[a + I*a*Tan[e +
f*x]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3590

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((A*b - a*B)*(a*c + b*d)*(a + b*Tan[e + f*x])^m)/(2*a^2*f*m), x] + Dist[
1/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[A*b*c + a*B*c + a*A*d + b*B*d + 2*a*B*d*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^{3/2}} \, dx &=\frac {(i A-B) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}-\frac {i \int \frac {a (B (c+i d)+A (i c+d))+2 a B d \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)}} \, dx}{2 a^2}\\ &=\frac {(i A-B) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {B (c+3 i d)+A (i c+d)}{2 a f \sqrt {a+i a \tan (e+f x)}}+\frac {((A-i B) (c-i d)) \int \sqrt {a+i a \tan (e+f x)} \, dx}{4 a^2}\\ &=\frac {(i A-B) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {B (c+3 i d)+A (i c+d)}{2 a f \sqrt {a+i a \tan (e+f x)}}-\frac {(i (A-i B) (c-i d)) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{2 a f}\\ &=-\frac {(i A+B) (c-i d) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(i A-B) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {B (c+3 i d)+A (i c+d)}{2 a f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 5.44, size = 206, normalized size = 1.40 \[ \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x)) \left (\frac {2}{3} \cos (e+f x) ((A (d+5 i c)+B (c+7 i d)) \cos (e+f x)-3 (A c-i A d-i B c+3 B d) \sin (e+f x))-i (A-i B) (c-i d) e^{i (e+f x)} \sqrt {1+e^{2 i (e+f x)}} \sinh ^{-1}\left (e^{i (e+f x)}\right )\right )}{4 f (a+i a \tan (e+f x))^{3/2} (A \cos (e+f x)+B \sin (e+f x)) (c \cos (e+f x)+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Tan[e + f*x])*(c + d*Tan[e + f*x]))/(a + I*a*Tan[e + f*x])^(3/2),x]

[Out]

(((-I)*(A - I*B)*(c - I*d)*E^(I*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcSinh[E^(I*(e + f*x))] + (2*Cos[e +
f*x]*((B*(c + (7*I)*d) + A*((5*I)*c + d))*Cos[e + f*x] - 3*(A*c - I*B*c - I*A*d + 3*B*d)*Sin[e + f*x]))/3)*(A
+ B*Tan[e + f*x])*(c + d*Tan[e + f*x]))/(4*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(c*Cos[e + f*x] + d*Sin[e + f*
x])*(a + I*a*Tan[e + f*x])^(3/2))

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fricas [B]  time = 0.82, size = 625, normalized size = 4.25 \[ -\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} c^{2} - {\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} c d - {\left (A^{2} - 2 i \, A B - B^{2}\right )} d^{2}}{a^{3} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (4 i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + 4 i \, a^{2} f\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} c^{2} - {\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} c d - {\left (A^{2} - 2 i \, A B - B^{2}\right )} d^{2}}{a^{3} f^{2}}} - {\left (4 \, {\left (A - i \, B\right )} a c - {\left (4 i \, A + 4 \, B\right )} a d\right )} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{{\left (A - i \, B\right )} c - {\left (i \, A + B\right )} d}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} c^{2} - {\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} c d - {\left (A^{2} - 2 i \, A B - B^{2}\right )} d^{2}}{a^{3} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-4 i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - 4 i \, a^{2} f\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} c^{2} - {\left (2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}\right )} c d - {\left (A^{2} - 2 i \, A B - B^{2}\right )} d^{2}}{a^{3} f^{2}}} - {\left (4 \, {\left (A - i \, B\right )} a c - {\left (4 i \, A + 4 \, B\right )} a d\right )} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{{\left (A - i \, B\right )} c - {\left (i \, A + B\right )} d}\right ) - \sqrt {2} {\left ({\left (i \, A - B\right )} c - {\left (A + i \, B\right )} d + {\left ({\left (4 i \, A + 2 \, B\right )} c + 2 \, {\left (A + 4 i \, B\right )} d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left ({\left (5 i \, A + B\right )} c + {\left (A + 7 i \, B\right )} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{12 \, a^{2} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/12*(3*sqrt(1/2)*a^2*f*sqrt(-((A^2 - 2*I*A*B - B^2)*c^2 - (2*I*A^2 + 4*A*B - 2*I*B^2)*c*d - (A^2 - 2*I*A*B -
B^2)*d^2)/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(-(sqrt(2)*sqrt(1/2)*(4*I*a^2*f*e^(2*I*f*x + 2*I*e) + 4*I*a^2*f)*
sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-((A^2 - 2*I*A*B - B^2)*c^2 - (2*I*A^2 + 4*A*B - 2*I*B^2)*c*d - (A^2 -
2*I*A*B - B^2)*d^2)/(a^3*f^2)) - (4*(A - I*B)*a*c - (4*I*A + 4*B)*a*d)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)/((A -
I*B)*c - (I*A + B)*d)) - 3*sqrt(1/2)*a^2*f*sqrt(-((A^2 - 2*I*A*B - B^2)*c^2 - (2*I*A^2 + 4*A*B - 2*I*B^2)*c*d
- (A^2 - 2*I*A*B - B^2)*d^2)/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(-(sqrt(2)*sqrt(1/2)*(-4*I*a^2*f*e^(2*I*f*x +
2*I*e) - 4*I*a^2*f)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-((A^2 - 2*I*A*B - B^2)*c^2 - (2*I*A^2 + 4*A*B - 2*
I*B^2)*c*d - (A^2 - 2*I*A*B - B^2)*d^2)/(a^3*f^2)) - (4*(A - I*B)*a*c - (4*I*A + 4*B)*a*d)*e^(I*f*x + I*e))*e^
(-I*f*x - I*e)/((A - I*B)*c - (I*A + B)*d)) - sqrt(2)*((I*A - B)*c - (A + I*B)*d + ((4*I*A + 2*B)*c + 2*(A + 4
*I*B)*d)*e^(4*I*f*x + 4*I*e) + ((5*I*A + B)*c + (A + 7*I*B)*d)*e^(2*I*f*x + 2*I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e
) + 1)))*e^(-3*I*f*x - 3*I*e)/(a^2*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (d \tan \left (f x + e\right ) + c\right )}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(d*tan(f*x + e) + c)/(I*a*tan(f*x + e) + a)^(3/2), x)

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maple [A]  time = 0.27, size = 131, normalized size = 0.89 \[ -\frac {2 i \left (-\frac {\left (\frac {1}{4} i A d +\frac {1}{4} i B c -\frac {1}{4} c A +\frac {1}{4} B d \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {-\frac {1}{4} i A d -\frac {1}{4} i B c +\frac {1}{4} c A +\frac {3}{4} B d}{\sqrt {a +i a \tan \left (f x +e \right )}}-\frac {a \left (i A d +i B c +c A -B d \right )}{6 \left (a +i a \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))*(c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2),x)

[Out]

-2*I/f/a*(-1/2*(1/4*I*A*d+1/4*I*B*c-1/4*c*A+1/4*B*d)*2^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan(f*x+e))^(1/2)*2^(1
/2)/a^(1/2))-(-1/4*I*A*d-1/4*I*B*c+1/4*c*A+3/4*B*d)/(a+I*a*tan(f*x+e))^(1/2)-1/6*a*(-B*d+I*A*d+I*B*c+c*A)/(a+I
*a*tan(f*x+e))^(3/2))

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maxima [A]  time = 0.64, size = 143, normalized size = 0.97 \[ \frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (A - i \, B\right )} {\left (c - i \, d\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (f x + e\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (f x + e\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (2 \, {\left (A + i \, B\right )} a c - {\left (-2 i \, A + 2 \, B\right )} a d + {\left (3 \, {\left (A - i \, B\right )} c - {\left (3 i \, A - 9 \, B\right )} d\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}\right )}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\right )}}{24 \, a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

1/24*I*(3*sqrt(2)*(A - I*B)*(c - I*d)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(f*x + e) + a))/(sqrt(2)*sqrt(a) + s
qrt(I*a*tan(f*x + e) + a)))/sqrt(a) + 4*(2*(A + I*B)*a*c - (-2*I*A + 2*B)*a*d + (3*(A - I*B)*c - (3*I*A - 9*B)
*d)*(I*a*tan(f*x + e) + a))/(I*a*tan(f*x + e) + a)^(3/2))/(a*f)

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mupad [B]  time = 10.27, size = 245, normalized size = 1.67 \[ \frac {\frac {\left (A\,c+A\,d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3\,f}+\frac {\left (A\,c-A\,d\,1{}\mathrm {i}\right )\,\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a\,f}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\frac {B\,c+B\,d\,1{}\mathrm {i}}{3\,f}-\frac {\left (B\,c+B\,d\,3{}\mathrm {i}\right )\,\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{2\,a\,f}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,B\,\left (d+c\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}\,\left (B\,c-B\,d\,1{}\mathrm {i}\right )}\right )\,\left (d+c\,1{}\mathrm {i}\right )}{4\,{\left (-a\right )}^{3/2}\,f}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,A\,\left (d+c\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}\,\left (A\,c-A\,d\,1{}\mathrm {i}\right )}\right )\,\left (d+c\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a^{3/2}\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(c + d*tan(e + f*x)))/(a + a*tan(e + f*x)*1i)^(3/2),x)

[Out]

(((A*c + A*d*1i)*1i)/(3*f) + ((A*c - A*d*1i)*(a + a*tan(e + f*x)*1i)*1i)/(2*a*f))/(a + a*tan(e + f*x)*1i)^(3/2
) - ((B*c + B*d*1i)/(3*f) - ((B*c + B*d*3i)*(a + a*tan(e + f*x)*1i))/(2*a*f))/(a + a*tan(e + f*x)*1i)^(3/2) +
(2^(1/2)*B*atanh((2^(1/2)*B*(c*1i + d)*(a + a*tan(e + f*x)*1i)^(1/2))/(2*(-a)^(1/2)*(B*c - B*d*1i)))*(c*1i + d
))/(4*(-a)^(3/2)*f) + (2^(1/2)*A*atan((2^(1/2)*A*(c*1i + d)*(a + a*tan(e + f*x)*1i)^(1/2))/(2*a^(1/2)*(A*c - A
*d*1i)))*(c*1i + d)*1i)/(4*a^(3/2)*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (e + f x \right )}\right ) \left (c + d \tan {\left (e + f x \right )}\right )}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c+d*tan(f*x+e))/(a+I*a*tan(f*x+e))**(3/2),x)

[Out]

Integral((A + B*tan(e + f*x))*(c + d*tan(e + f*x))/(I*a*(tan(e + f*x) - I))**(3/2), x)

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